Q4
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└── Homework
└── W4
└── Q4.tex
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\begin{document}
\section*{Question 4}
\noindent Consider a routine screening test for a disease. Suppose the frequency of the disease in the population (base rate) is $0.5\%$. The test is fairly accurate with a $5\%$ false positive rate and a $10\%$ false negative rate. You take the test and it comes back positive. What is the probability that you have the disease?
\bigskip
\begin{enumerate}[label={},leftmargin=0in]\item
\subsection*{Solution}
Let $\it{positive}$ and $\it{negative}$ denote the results of a test. Define the probability space as
\[
\begin{aligned}
\Omega &= \{\mathrm{positive},\,\mathrm{negative}\}\\
\mathcal{F} &= \mathcal{P}(\Omega)\\
\bP &:\enspace \text{the probability measure on $\mathcal{F}$}
\end{aligned}
\]
Let $B$ be the event that having the disease in the population (base rate)
\[\bP(B) = 0.5\%\]
Then we have the false positive rate ($\bP_{B^c}(\{\mathrm{positive}\})$) and the false negative rate ($\bP_B(\{\mathrm{negative}\})$)
\[
\begin{aligned}
\bP_{B^c}(\{\mathrm{positive}\}) &= 5\%\\
\bP_B(\{\mathrm{negative}\}) &= 10\%
\end{aligned}
\]
Then we have the true negative rate ($\bP_{B^c}(\{\mathrm{negative}\})$) and the true positive rate ($\bP_B(\{\mathrm{positive}\})$)
\[
\begin{aligned}
\bP_{B^c}(\{\mathrm{negative}\}) &= 1-\bP_{B^c}(\{\mathrm{positive}\}) &= 95\%\\
\bP_B(\{\mathrm{positive}\}) &= 1-\bP_B(\{\mathrm{negative}\}) &= 90\%
\end{aligned}
\]
Then by applying the law of total probability, we have
\[\bP(\{\mathrm{positive}\}) = \bP(B)\bP_B(\{\mathrm{positive}\}) + \bP(B^c)\bP_{B^c}(\{\mathrm{positive}\}) = 5.425\%\]
By applying the Bayes’ Theorem, we have
\[
\bP_{\{\mathrm{positive}\}}(B) = \frac{\bP_B(\{\mathrm{positive}\})\bP(B)}{\bP(\{\mathrm{positive}\})} = \frac{18}{217} \approx 8.295\%
\]
\subsection*{Answer}
\[\boxed{\bP_{\{\mathrm{positive}\}}(B) = \frac{18}{217} \approx 8.295\%}\]
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